原题传送门
发现要是两个数
,其实只需要关注他俩是否有质数公因数
对于每个数暴力分解质因数,树形dp
表示点
为根,第
个质因数的最长链
Code:
#include <bits/stdc++.h>
#define maxn 200010
using namespace std;
vector <int> prime[maxn], dp[maxn];
struct Edge{
int to, next;
}edge[maxn << 1];
int num, head[maxn], n, ans;
inline int read(){
int s = 0, w = 1;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
return s * w;
}
void addedge(int x, int y){ edge[++num] = (Edge){y, head[x]}, head[x] = num; }
void dfs(int u, int pre){
for (int i = head[u]; i; i = edge[i].next){
int v = edge[i].to;
if (v != pre){
dfs(v, u);
for (int j = 0; j < prime[u].size(); ++j)
for (int k = 0; k < prime[v].size(); ++k)
if (prime[u][j] == prime[v][k])
ans = max(ans, dp[u][j] + dp[v][k]), dp[u][j] = max(dp[u][j], dp[v][k] + 1);
}
}
}
int main(){
n = read();
for (int i = 1; i <= n; ++i){
int x = read();
if (x > 1) ans = 1;
for (int j = 2; j * j <= x; ++j)
if (x % j == 0){
prime[i].push_back(j);
dp[i].push_back(1);
while (x % j == 0) x /= j;
}
if (x > 1) prime[i].push_back(x), dp[i].push_back(1);
}
for (int i = 1; i < n; ++i){
int x = read(), y = read();
addedge(x, y), addedge(y, x);
}
dfs(1, 0);
printf("%d\n", ans);
return 0;
}